Determining the Number of Oxygen Atoms in Compounds: Percent Composition and Molar Ratios, Slides of Law

Download Determining the Number of Oxygen Atoms in Compounds: Percent Composition and Molar Ratios and more Slides Law in PDF only on Docsity! How do we figure this out? We know that: 1) the number of oxygen atoms can be found by using Avogadro’s number, if we know the moles of oxygen atoms; 2) the number of moles of oxygen atoms can be found if we know the number of moles of each molecule; 3) the number of moles of each molecule can be found if we know the mass of each compound We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do? Since we have an equal mass of each compound, if we calculate the percent by mass of oxygen in each compound, we could just compare their ratios, since we know that the mass is directly related to number of moles (which is directly related to the number of atoms). So let’s start by finding the % by mass of oxygen in H2SO4: To find % mass, we use the equation: As we did with the first example we worked in class, we assume we have 1 mole of the compound, which has a molar mass of 98.086 g/mol (mass of the whole): For every mole of H2SO4 molecules, there are 4 moles of oxygen atoms, which has a molar mass of 16.00 g/mol (mass of the part): Now we just punch in the numbers and find: Now we can do the same for the other two compounds! Assuming 1 mole of sugar (mass of the whole: 342.296g/mol), we have 11 moles of oxygen atoms per mole of sugar molecules: Now I have moles, but since it is moles of the compound, not just oxygen. I’ll need to convert the moles of the compound to moles of oxygen atoms: Well, I have moles of oxygen atoms, so I can figure out the number of moles of oxygen atoms with Avogadro’s number, which is number of “things” per mole: So now all of my units should cancel out correctly, leaving me with the number of oxygen atoms! So all I have to do is plug in the numbers and solve… Note that the starting amount given only has three significant figures, so your answer should have the same. Now we can do the same calculation for the other two compounds: For sugar (C12H22041): vs CAP, ue BACHIP yg [val Bows], CPIArIO“ohong — 50.0 ° priosy — (wl Ep, ml fon 0 61 AP? S| I 61x aor O For potassium chlorate (KCIOs): 50.04 KCI, Heb KCIOr Seal stoes , GDparI0" oho 2 ° D.5515 — Twel KEIO, ral ent axle Slo sq a | 0a O  So we want to find the formula and name of some iron oxide compound. What do we do? Well, we know that the new compound is some “iron oxide” meaning that it must be composed of X moles of iron and Y moles of oxygen: FexOy So if we want to write a formula, we need to figure out the molar ratios of iron and oxygen in the new compound so we can figure out the subscripts. But how can we even know how much of each element there is in the new compound? Remember that we learned about the Law of Conservation of Mass earlier this semester, which means that whatever amount of matter we start with before the reaction will be the amount of matter we end up with after the reaction. Since we start with 12.00 grams of iron, that means that the new compound contains 12.00 g of iron as well (assuming all of the iron is in the new compound now). So the moles of iron in the new compound should be: So here we have another problem in which we have to find the molecular formula, but this time we do not know how much we started with, or how much we ended up with, we are only given the mass of carbon in propane and its molar mass. So what can’t do what we did on the previous problem. What can we do? Another way we can figure out ratios of substances in a compound is to find their percent composition by mass. First, we’ll start by calculating the % by mass of carbon, since we are given the amount of carbon (mass of part) and the molar mass (mass of whole): If propane is 81.71% carbon, then the % hydrogen must be: In order to find the molecular formula of propane, though, we need to find the moles of each particle to figure out the ratios. As I talked about in lecture, percent composition is independent of the amount of a substance. So we can choose any amount to calculate the moles. And since percents are out of 100%, it’s easiest to just set our amount to 100 grams. So if we have 100 grams of propane, then there are 81.71 grams of carbon and 18.29 grams of hydrogen. Using molar masses, we can find moles of each: Then we find the ratio of the two: And finally we find the closest whole number ratio: So we have found the formula: C3H8 But is this the empirical formula or the molecular formula? Well, we know the molar mass of propane should be 44.094 g/mol, so let’s check what the molar mass of C3H8 would be: It checks out! So C3H8 is the molecular formula, but since it is the smallest whole number ratio, it is also the empirical formula! This problem is very similar to the one we just did, but this problem was really nice and gave us two of the three % compositions we need! To find % by mass of oxygen, we just have to subtract the % by mass of carbon and hydrogen from 100%: Now that we have all three % compositions, we can find the moles of each particle as we did before by assuming that we have 100 grams of the compound, which means we have: 40.00 grams of carbon, 6.71 grams of hydrogen and 53.29 grams of oxygen. Using molar mass to find moles, we get: